Question

Complete Example 2 by showing that

${\int }_0^{\pi /3}{\int }_0^{1+\cos \theta }rdrd\theta =\frac{1}{4}\pi +\frac{9}{16}\sqrt{3}$

and

${\int }_{\pi /3}^{\pi /2}{\int }_0^{3\cos \theta }rdrd\theta =\frac{3}{8}\pi -\frac{9}{16}\sqrt{3}$

Short answer

The required proof is${\int }_{\pi /3}^{\pi /2}{\int }_0^{los\theta }rdrd\theta =\frac{3\pi }{8}-\frac{9}{16}\sqrt{3}$

Step-by-step solution

Given Information

The objective is to show that,

${\int }_0^{\pi /3}{\int }_0^{1+ses\theta }rdrd\theta =\frac{1}{4}\pi +\frac{9}{16}\sqrt{3}$

and

${\int }_{\pi /3}^{\pi /2}{\int }_0^{3\cos \theta }rdrd\theta =\frac{3}{8}\pi -\frac{9}{16}\sqrt{3}$

Calculation

${\int }_0^{\pi /3}{\int }_0^{1+\cos \theta }rdrd\theta$

Integrate with respect to $r$.

${\int }_0^{\pi /3}{\int }_0^{1+ces\theta }rdrd\theta ={\int }_0^{\pi /3}{\left[\frac{r^2}{2}\right]}_0^{1+\cos \theta }d\theta$

Substitute the limits

${\int }_0^{\pi /3}{\int }_0^{1+ses\theta }rdrd\theta ={\int }_0^{\pi /3}\left[\frac{(1+\cos \theta {)}^2-0}{2}\right]d\theta$

${\int }_0^{\pi /3}{\int }_0^{1+\cos \theta }rdrd\theta ={\int }_0^{\pi /3}\left[\frac{\left(1+2\cos \theta +{\cos }^2\theta \right)}{2}\right]d\theta \left[(a+b{)}^2=a^2+2ab+b^2\right]$

${\int }_0^{\pi /3}{\int }_0^{1+\cos \theta }rdrd\theta ={\int }_0^{\pi /3}\left[\frac{\{1+2\cos \theta +(1+\cos 2\theta )/2\}}{2}\right]\theta$

${\int }_0^{\pi /3}{\int }_0^{1+\cos \theta }rdrd\theta ={\int }_0^{\pi /3}\left[\frac{3+4\cos \theta +\cos 2\theta }{4}\right]d\theta$

Integrate with respect to $\theta$.

${\int }_0^{\pi /3}{\int }_0^{1+\cos \theta }rdrd\theta ={\left[\frac{3\theta +4\sin \theta +\left(\sin 2\theta \right)/2}{4}\right]}_0^{*/3}\left[\int \cos xdx=\sin x\right]$

Put the limits

${\int }_0^{\pi /3}{\int }_0^{1+\cos \theta }rdrd\theta =\left[\frac{\pi +4\sin (\pi /3)+(\sin 2\pi /3)/2-0}{4}\right]$

${\int }_0^{z/3}{\int }_0^{1+ces\theta }rdrd\theta =\left[\frac{\pi +2\sqrt{3}+\sqrt{3}/4}{4}\right]$

${\int }_0^{\pi /3}{\int }_0^{1+\cos \theta }rdrd\theta =\frac{\pi }{4}+\frac{9}{16}\sqrt{3}$

Integration

${\int }_{\pi /3}^{\pi /2}{\int }_0^{3\cos \theta }rdrd\theta$

Integrate with respect to$r$

${\int }_{\pi /3}^{\pi /2}{\int }_0^{sen\theta }rdrd\theta ={\int }_{\pi /3}^{\pi /2}{\left[\frac{r^2}{2}\right]}_0^{3sen\theta }d\theta$

Substitute the limits

${\int }_{\pi /3}^{\pi /2}{\int }_0^{3\cos \theta }rdrd\theta ={\int }_{\pi /3}^{\pi /2}\left[\frac{(3\cos \theta {)}^2-0}{2}\right]d\theta$

${\int }_{\pi /3}^{\pi /2}{\int }_0^{3\cos \theta }rdrd\theta ={\int }_{\pi /3}^{\pi /2}\left[\frac{9{\cos }^2\theta }{2}\right]d\theta$

${\int }_{\pi /3}^{\pi /2}{\int }_0^{3\cos \theta }rdrd\theta =\frac{9}{4}{\int }_{\pi /3}^{\pi /2}[1+\cos 2\theta ]d\theta \left[{\cos }^2\theta =\frac{1}{2}(1+\cos 2\theta )\right]$

Integrate with respect to $\theta$.

${\int }_{\pi /3}^{\pi /2}{\int }_0^{3\cos \theta }rdrd\theta =\frac{9}{4}{\left[\theta +\frac{\sin 2\theta }{2}\right]}_{x/3}^{x/2}$

Substitute the limits

${\int }_{\pi /3}^{\pi /2}{\int }_0^{3\cos \theta }rdrd\theta =\frac{9}{4}\left[\left(\frac{\pi }{2}+\frac{\sin \pi }{2}\right)-\left(\frac{\pi }{3}+\frac{\sin (2\pi /3)}{2}\right)\right]$

${\int }_{\pi /3}^{\pi /2}{\int }_0^{3\cos \theta }rdrd\theta =\frac{9}{4}\left[\left(\frac{\pi }{2}\right)-\left(\frac{\pi }{3}+\frac{\sqrt{3}}{4}\right)\right]$

role="math" localid="1650520240502" ${\int }_{\pi /3}^{\pi /2}{\int }_0^{3\cos \theta }rdrd\theta =\frac{3\pi }{8}-\frac{9}{16}\sqrt{3}$

role="math" localid="1650520257012" ${\int }_{\pi /3}^{\pi /2}{\int }_0^{3c\mathrm{os}\theta }rdrd\theta =\frac{3\pi }{8}-\frac{9}{16}\sqrt{3}$